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Chapter 22: Problem 20

(II) A flat square sheet of thin aluminum foil, \(25 \mathrm{~cm}\) on a side, carries a uniformly distributed \(275 \mathrm{nC}\) charge. What, approximately, is the electric field (a) \(1.0 \mathrm{~cm}\) above the center of the sheet and (b) \(15 \mathrm{~m}\) above the center of the sheet?

Short Answer

Expert verified
(a) Approximately 248 N/C at 1 cm, (b) approximately 1.099 N/C at 15 m.

Step by step solution

01

Understand the Problem

We are given a square aluminum foil of side length 25 cm with a uniform charge distribution of 275 nC. We need to find the electric field at two different distances: 1 cm above and 15 m above the center of the sheet.
02

Convert Units

First, convert measurements to meters. The side length of the sheet is \( 25 \text{ cm} = 0.25 \text{ m} \). The charge is given as \( 275 \text{ nC} = 275 \times 10^{-9} \text{ C} \). The distances are \( 1.0 \text{ cm} = 0.01 \text{ m} \) and \( 15 \text{ m} \).
03

Calculate Surface Charge Density

Calculate the surface charge density \( \sigma \) using \( \sigma = \frac{Q}{A} \), where \( Q = 275 \times 10^{-9} \text{ C} \) and \( A = (0.25 \text{ m})^2 \). So, \( A = 0.0625 \text{ m}^2 \).Then, \( \sigma = \frac{275 \times 10^{-9} \text{ C}}{0.0625 \text{ m}^2} \approx 4.4 \times 10^{-9} \text{ C/m}^2 \).
04

Use Approximation for Planar Sheet for E at 1 cm

For small distances close to the sheet, assume the sheet acts like an infinite plane. The electric field due to an infinite plane is given by:\( E = \frac{\sigma}{2\varepsilon_0} \). Using \( \varepsilon_0 = 8.854 \times 10^{-12} \text{ C}^2/\text{N} \, \text{m}^2 \), calculate:\( E \approx \frac{4.4 \times 10^{-9}}{2 \times 8.854 \times 10^{-12}} \approx 248 \text{ N/C} \).
05

Consider Far Field Approximation for E at 15 m

For large distances from the sheet, treat the sheet as a point charge where \( E = \frac{kQ}{r^2} \).Calculate using \( k = 8.99 \times 10^9 \text{ N} \, \text{m}^2/\text{C}^2 \):\( E = \frac{8.99 \times 10^9 \times 275 \times 10^{-9}}{15^2} \approx 1.099 \text{ N/C} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Charge Distribution
Charge distribution refers to how electric charge is spread out on a given object. In our exercise, the charge is distributed uniformly over a square sheet of aluminum. This means every part of the sheet has the same amount of charge per unit area. When dealing with charge distribution:
  • A uniform distribution allows us to use simplified equations to calculate electric fields.
  • If the distribution were non-uniform, finding the electric field could become much more complicated.
For most physics problems at this level, assuming a uniform charge distribution makes calculations feasible and understandable. Visualize the charge as a thin blanket covering the sheet evenly. This blanket of charge influences the electric field around it, affecting nearby charges.
Surface Charge Density
Surface charge density, denoted as \( \sigma \), is the measure of how much charge is present per unit area on a surface. In this problem:
  • We calculated \( \sigma \) by dividing the total charge, \( Q = 275 \times 10^{-9} \text{ C} \), by the surface area, \( A = 0.0625 \text{ m}^2 \).
  • Thus, \( \sigma = \frac{275 \times 10^{-9} \text{ C}}{0.0625 \text{ m}^2} = 4.4 \times 10^{-9} \text{ C/m}^2 \).
Understanding surface charge density is crucial because it directly influences the electric field strength produced by the charged object. In essence, the higher the surface charge density, the stronger the electric field generated in the space around the charged surface.
Finite and Infinite Plane Approximation
In electrostatics, approximations help simplify complex calculations. Two common approximations are the finite and infinite plane approximations.
  • **Infinite Plane Approximation:** This assumes that a plane is large enough that its edges don't matter for nearby points. The electric field from such a plane is calculated using \( E = \frac{\sigma}{2\varepsilon_0} \). For this exercise, we used this to find the electric field 1 cm from the sheet, simplifying the calculations.
  • **Finite Plane Approximation:** Over longer distances, we can't ignore the boundaries of the plane. In such cases, treating the sheet as a point charge is more accurate. This is because, at large distances, the sheet's lateral dimensions become negligible compared to the distance studied.
These approximations are powerful tools in electrostatics, allowing us to make sense of complex electromagnetic interactions simply.
Electrostatics
Electrostatics is the branch of physics dealing with the study of forces, fields, and potentials arising from static (non-moving) charges. In our exercise, the electric field is central as it governs the force that a charged object can exert on other charges in its vicinity.Key points in electrostatics you should know:
  • **Electric Field (E):** A vector field around a charged object representing the force a positive test charge would experience. Calculated differently depending on proximity and distribution of source charges, as demonstrated in our exercise.
  • **Permittivity of Free Space (\( \varepsilon_0 \)):** A fundamental constant denoting the ability of a vacuum to permit electric field lines. It's crucial in formulas like \( E = \frac{\sigma}{2\varepsilon_0} \).
By understanding these core concepts, students can better grasp how charges interact with each other in space, which is vital in both theoretical studies and practical applications.

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Most popular questions from this chapter

(III) A flat slab of nonconducting material (Fig. \(22-40\) ) carries a uniform charge per unit volume, \(\rho_{\mathrm{E}}\). The slab has thickness \(d\) which is small compared to the height and breadth of the slab. Determine the electric field as a function of \(x(a)\) inside the slab and ( \(b\) ) outside the slab (at distances much less than the slab's height or breadth). Take the origin at the center of the slab.

(II) A very long solid nonconducting cylinder of radius \(R_{1}\) is uniformly charged with a charge density \(\rho_{\mathrm{E}}\). It is surrounded by a concentric cylindrical tube of inner radius \(R_{2}\) and outer radius \(R_{3}\) as shown in Fig. \(22-36,\) and it too carries a uniform charge density \(\rho_{\mathrm{E}}\). Determine the electric field as a function of the distance \(R\) from the center of the cylinders for \((a) 0R_{3}\). \((e)\) If \(\rho_{\mathrm{E}}=15 \mu \mathrm{C} / \mathrm{m}^{3}\) and \(R_{1}=\frac{1}{2} R_{2}=\frac{1}{3} R_{3}=5.0 \mathrm{~cm}\) plot \(E\) as a function of \(R\) from \(R=0\) to \(R=20.0 \mathrm{~cm} .\) Assume the cylinders are very long compared to \(R_{3}\)

(II) A flat ring (inner radius \(R_{0},\) outer radius 4\(R_{0} )\) is uniformly charged. In terms of the total charge \(Q,\) determine the electric field on the axis at points \((a) 0.25 R_{0}\) and (b) 75\(R_{0}\) from the center of the ring. [Hint: The ring can be replaced with two oppositely charged superposed disks.]

(II) A thin cylindrical shell of radius \(R_{1}=6.5 \mathrm{~cm}\) is surrounded by a second cylindrical shell of radius \(R_{2}=9.0 \mathrm{~cm},\) as in Fig. \(22-35 .\) Both cylinders are \(5.0 \mathrm{~m}\) long and the inner one carries a total charge \(Q_{1}=-0.88 \mu \mathrm{C}\) and the outer one \(Q_{2}=+1.56 \mu \mathrm{C}\). For points far from the ends of the cylinders, determine the electric field at a radial distance \(r\) from the central axis of \((a) 3.0 \mathrm{~cm},(b) 7.0 \mathrm{~cm},\) and \((c) 12.0 \mathrm{~cm} .\)

(III) Charge is distributed within a solid sphere of radius \(r_{0}\) in such a way that the charge density is a function of the radial position within the sphere of the form: \(\rho_{\mathrm{E}}(r)=\rho_{0}\left(r / r_{0}\right) .\) If the total charge within the sphere is \(Q\) (and positive), what is the electric field everywhere within the sphere in terms of \(Q, r_{0},\) and the radial position \(r ?\)
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